Adventures in Technology and Science.

Scientific Sermon: Parity Game Solving

[latexpage]Last time, we saw how parity games look like and what their complexity theoretic status is – we can solve them by a non-deterministic polytime algorithm, but we don’t know how to do it in deterministic polytime.

In this post, we will see one of the most natural algorithms for solving parity games deterministically. Along the way, we will see why parity games are positionally determined, i.e. why every node is won by one of the two players by using a positional strategy, no matter how the opponent is playing. Recall that a general strategy is a plan that tells the player which outgoing edge to take from a node by taking the whole history of the play into account. Positional strategies, on the other hand, don’t care about the history of the play. In other words, no matter how the history looks like, a positional strategy will always choose the same outgoing edge from a node.

Before we get started, let us fix a mathematical formulation of parity games first. We say that $G=(V,V_0,V_1,E,\Omega)$ is a parity game if $(V,E)$ is a directed, total graph – i.e. the arena on which the game is based -, $V_0$ and $V_1$ are a partition of $V$ into the node sets of each player – if you recall the last post, the circular nodes belong to $V_0$ while the rectangular posts belong to $V_1$ -, and $\Omega: V \rightarrow \mathbb{N}$ is the priority assignment function that maps each node to a natural number, its priority.

The algorithm is based on the notion of attractors. Let $i=0,1$ be one of the two players. Given a set of nodes $U$, the $i$-attractor of $U$ is the least superset $A$ of $U$ such that player $i$ can force every play starting in a node $v \in A$ to visit a node $u \in U$. Intuitively, you can think of $U$ as a set of nodes beneficial to player $i$ and $A$ to be an extension in which every path, no matter what the other player is doing, leads to the set $U$.

Mathematically, the attractor $A$ is determined iteratively, starting with $A := U$. In every step, we include all nodes by player $i$ that have an edge leading directly to $A$. Obviously, player $i$ can force these nodes to visit $A$ and by induction to visit $U$. Also, we include all nodes by the other player that have all outgoing edges directed towards $A$. In other words, no matter what player $1-i$ is doing, he ends up in $A$ and by induction in $U$. This process is iterated until no new nodes have been added to $A$. It is immediate to see that the attractor computation can be done in deterministic polytime. Let $Attr_i(U)$ denote the $i$-attractor of $U$.

It should be easy to see that player $i$ can force a visit from $A$ to $U$ by a positional strategy.

The algorithm for solving parity games that we consider here is called the recursive algorithm for parity games. It is based on a recursive descent in the number of priorities. The base case is the empty game. Assume now that the game $G$ is not empty and let $p$ be the largest priority occurring. Let $i$ be the player “owning” the priority, i.e. $i = p \mod 2$. Let $U$ be the set of nodes with priority $p$. It is easy to see that every play that visits $U$ infinitely often is won by player $i$, because the largest priority occurring is $p$. So consider $U$ to be a “good” region for player $i$. We can even extend $U$ to a potentially larger good region by building the $i$-attractor $A = Attr_i(U)$. Every play that visits $A$ infinitely often is won by player $i$.

The question remains whether $i$ can enforce infinitely many visits to $A$. In order to answer that question, we consider the game $G’ = G \setminus A$ in which all $A$-nodes and $A$-edges have been removed. Note that the arena $G’$ is total again only due to the fact that $A$ is an attractor. By recursion, we solve $G’$ and obtain a partition into winning sets for both players $W_i’$ and $W_{1-i}’$.

There are two cases now. The easy one is $W_{1-i}’ = \emptyset$, i.e. player $1-i$ can’t win anything in the sub game G’. In that case, player $i$ wins from every node in $G$, because if player $1-i$ decides to visit $A$ infinitely often, player $i$ wins by forcing the visits to $U$, and if player $1-i$ decides to visit $A$ only finitely often, player $i$ wins by induction. It is easy to see by induction that this can be done with a positional strategy.

The other case is $W_{1-i}’ \not= \emptyset$. Here, we only we only know for sure that player $1-i$ can win on $W_{1-i}’$ as there is no player $i$ escape from $W_{1-i}’$ to $A$ (because $A$ is an attractor already). So what can we do now? We enlarge $W_{1-i}’$ again by computing the $1-i$-attractor $B = Attr_{1-i}(W_{1-i}’)$. It is immediate to see that player $1-i$ wins on $B$ with a positional strategy. We consider a second sub game $G” = G \setminus B$ and solve it recursively again. We obtain a partition into winning sets $W_i”$ and $W_{1-i}”$, which we can use to define the winning sets for $G$, i.e. $W_i := W_i”$ and $W_{1-i} := W_{1-i}” \cup B$.

I’ve updated the respective Wikipedia article as well where you can find a pseudocode outline of the algorithm.

The recursive algorithm is probably the most straight-forward algorithm for solving parity games. It can be easily seen that the algorithm is potentially exponential because of the two recursive descents and one can, in fact, provide a family of games on which the algorithm runs in exponential time. Interestingly, the algorithm is one of the fastest in practice. It also shows that all nodes of a parity game can be partitioned into the winning sets of both players and that they can win with a positional strategy on their respective winning sets.

The next post in this series will consider some interesting, seemingly simpler subclasses of parity games that would allow us to define a polytime algorithm for general parity games, given that we would have a polytime algorithm for one of the simpler classes.